2. What is the strain energy stored in a body when the load is applied suddenly?
a) σE/V
b) σE2/V
c) σV2/E
d) σV2/2E

3. A tensile load of 60kN is suddenly applied to a circular bar of 4cm diameter. What will be the maximum instantaneous stress induced?
a) 95.493 N/mm2
b) 45.25 N/mm2
c) 85.64 N/mm2
d) 102.45 N/mm2

Explanation: Maximum instantaneous stress induced = 2P/A = 2×60000/400π = 95.49 N/mm2.

4. A tensile load of 60kN is suddenly applied to a circular bar of 4cm and 5m length. What will be the strain energy absorbed by the rod if E=2×105 N/mm2?
a) 140.5 N-m
b) 100 N-m
c) 197.45 N-m
d) 143.2 N-m

Explanation: Maximum instantaneous stress induced = 2P/A = 2×60000/400π = 95.49 N/mm2
Strain energy = σ2V/2E = 95.492 x 2×106π / (2x2x105) = 143238 N-mm = 143.23 N-m.

5. A tensile load of 100kN is suddenly applied to a rectangular bar of dimension 2cmx4cm. What will be the instantaneous stress in bar?
a) 100 N/mm2
b) 120 N/mm2
c) 150 N/mm2
d) 250 N/mm2

Explanation: Stress = 2x load / area = 2×100,000/ (20×40) = 250 N/mm2.
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6. 2 tensile load of 100kN is suddenly applied to a rectangular bar of dimension 2cmx4cm and length of 5m. What will be the strain energy absorbed in the bar if E=1×105 N/mm2?
a) 312.5 N-m
b) 314500 N-mm
c) 1250 N-m
d) 634 N-m

Explanation: Stress = 2xload / area = 2×100,000/ (20×40) = 250 N/mm2
Strain energy = σ2V/2E = 250x250x20x40x5000/ (2×100,000) = 1250000 N-mm = 1250 N-m.

7. A steel rod is 2m long and 50mm in diameter. A axial pull of 100kN is suddenly applied to the rod. What will be the instantaneous stress induced in the rod?
a) 101.89 N/mm2
b) 94.25 N/mm2
c) 130.45 N/mm2
d) 178.63 N/mm2

Explanation: Area = π/4 d2 = 625π
Load = 100kN = 100×1000 N
Stress = 2 x load / area = 2x100x1000 / (625π) = 101.86 N/mm2.

8. A steel rod is 2m long and 50mm in diameter. An axial pull of 100kN is suddenly applied to the rod. What will be the instantaneous elongation produced in the rod if E=22GN/m2?
a) 0.0097 mm
b) 1.0754 mm
c) 1.6354 mm
d) 1.0186 mm

Explanation: Area = π/4 d2 = 625π
Load = 100kN = 100×1000 N
E=22GN/m2 = 200 x 109 / 106 = 200,000 N/mm2
Stress = 2 x load / area = 2x100x1000 / (625 π )
Elongation = stress x length / E = 101.86×2000 / 200000 = 1.0186 mm.

9. What will be the amount of axial pull be applied on a a 4cm diameter bar to get an instantaneous stress value of 143 N/mm2?
a) 50kN
b) 60kN
c) 70kN
d) 80kN

Explanation: Instantaneous stress = 2 x load / area
Load = instantaneous stress x area / 2
= 143 x 400×3.14 / 2 = 60kN.

10. What will be the instantaneous stress produced in a bar 10cm2 in area ans 4m long by the sudden application of tensile load of unknown magnitude, if the extension of the bar due to suddenly applied load is 1.35mm if E = 2×105 N/mm2?
a) 67.5 N/mm2
b) 47 N/mm2
c) 55.4 N/mm2
d) 78.5 N/mm2

Explanation: The value of stress = load / area where area is 10cm2 and load can be calculated by stress strain equation.

Sanfoundry Global Education & Learning Series – Strength of Materials.

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